Starting from the point A - 3 , 4 , a moving object touches 2 x + y - 7 = 0 at B and reaches the point C 0 ,…

Given,

Point $A\left(-3,4\right)$ and $C\left(0,1\right)$ and let moving point be $B\left(x,y\right)$

Now given equation of line $2x+y=7\Rightarrow y=-2x+7 ...\left(1\right)$

Now finding distance we get, $AB+BC=\sqrt{{\left(x+3\right)}^2+{\left(y-4\right)}^2}+\sqrt{{\left(x-0\right)}^2+{\left(y-1\right)}^2} ...\left(2\right)$

Now substituting the value of $y$ in (2) we get,

$=\sqrt{{\left(x+3\right)}^2+{\left(-2x+7-4\right)}^2}+\sqrt{{\left(x-0\right)}^2+{\left(-2x+7-1\right)}^2}$

$=\sqrt{x^2+9+6x+4x^2+9-12x}+\sqrt{x^2+4x^3+36-24x}$

$=\sqrt{5x^2-6x+18}+\sqrt{5x^2-24x+36}$

Now let $f\left(x\right)=\sqrt{5x^2-6x+18}+\sqrt{5x^2-24x+36}$

Now to get minimum value we will differentiate wrt $x$ we get,

$f^{\text{'}}\left(x\right)=\frac{10x-6}{2\sqrt{5x^2-6x+18}}+\frac{10x-24}{2\sqrt{5x^2-24x+36}}$

$\Rightarrow f^{\text{'}}\left(x\right)=\frac{5x-3}{\sqrt{5x^2-6x+18}}+\frac{5x-12}{\sqrt{5x^2-24x+36}}=0$

$\Rightarrow \frac{5x-3}{\sqrt{5x^2-6x+18}}=\frac{12-5x}{\sqrt{5x^2-24x+36}}$

Now squaring on bothisdes we get,

$\Rightarrow \frac{{\left(5x-3\right)}^2}{5x^2-6x+18}=\frac{{\left(12-5x\right)}^2}{5x^3-24x+36}$

$\Rightarrow \frac{25x^2+9-30x}{5x^2-6x+18}=\frac{144+25x^2-120x}{5x^2-24x+36}$

By solving the above equation we get $25x^2-192x+252=0$

As it is a quadratic equation, we get two values of $x$.

So, $x=\frac{+192\pm \sqrt{{\left(192\right)}^2-4\left(252\right)\left(25\right)}}{2\left(25\right)}$

$\Rightarrow x=\frac{192\pm \sqrt{11664}}{50}$

$\Rightarrow x=\frac{192\pm 108}{50}$

So, $x= \frac{192+108}{50} \text{or} \frac{192-108}{50}$

$\Rightarrow x=6 \text{or} \frac{42}{25}$

So we will take $x=\frac{42}{25}$ as it is a smaller value and we want minimum distance,

Then by substitute $x$ in the equation we get $y=\frac{91}{25}$

So, $B\left(x,y\right)=\left(\frac{42}{25},\frac{91}{25}\right)$

Now by distance formula we get $AB=\frac{9\sqrt{170}}{25}$